Prove that if a triangle's area and sides are all integers, its perimeter must be even.

let A be the area

a,b,c be the sides

p be the perimeter

s=p/2

then

A^2=s(s-a)(s-b)(s-c)

A^2=p(p-2a)(p-2b)(p-2c)/16

16*A^2=p(p-2a)(p-2b)(p-2c)

now if p is odd then p=2k+1 for integer k

and

16*A^2=(2k+1)(2(p-a)+1)(2(p-b)+1)(2(p-c)+1)

left side is even but right side has all odd factors and thus

can not be even so therefore p can not be odd and thus must be even.

Posted by Daniel on 2009-01-16 12:26:51